Elitmus
Exam
Numerical Ability
LCM and HCF
The addition 742586 + 829430 = 1212016 is incorrect. It can be corrected by changing
one digit d, wherever it occurs, to another digit e. Find sum of d and e.
A. 6 [DCRU]
B. 4
C. 10
D. 8
Read Solution (Total 7)
-
- To solve this, basically you have to try hit and trial, using options, and as changing 2 digit is most suitable to get answer, let's choose 8=2+6. So,
Consider d=2 and e=6
Therefore, changing every occurrence of 2 with 6 in the whole statement(i.e both R.H.S and L.H.S), we get
746586+869430=1616016 - 9 years agoHelpfull: Yes(10) No(2)
- There's some problem with question . Although i have tried and here it is :
if we change 7 to 3 in number 1 and 2 to 6 in second number the number would become 342586 and 869430 and the sum will be 1212016 . so here d could be 6 and e could be 2 so d+e =8.
and we can get d = 3 and e=7 and sum will be d+e =10 .There's a conflict between options . Hope y'all get it. - 9 years agoHelpfull: Yes(7) No(0)
- 742586+829430=1272016,
incorrect ans is 1212016, we can change 1 by 7 to get correct ans,
so 7+1=8(ans) - 9 years agoHelpfull: Yes(1) No(3)
- After analysing every digit properly if you replace every '2' by '6' you will get a new answer which will be appropriate. Hence here d=2 and e=6.
- 8 years agoHelpfull: Yes(1) No(0)
- no clarity in question... what it mean "wherever it occurs, to another digit e"?
- 9 years agoHelpfull: Yes(0) No(0)
- sum of 742586 and 829430 =1572016
but given sum is =1212016
so if we consider d as 7 and the digit to be replaced i.e 1
then option d
i am not sure
let me know - 9 years agoHelpfull: Yes(0) No(1)
- 742586+829430= 1212016 is incorrect. but when we replace 2 by 6 and 8 by 4 then we got correct answer.
742586+469430=1212016
d=6
e=4
d+e=6+4=10 - 4 years agoHelpfull: Yes(0) No(0)
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